5
$\begingroup$

I am supposed to show that if $f$ and $g$ are continuous, real-valued functions on $\mathbb{R}$, then if $f=g\;\;$, $\lambda$-a.e., then $f=g$ everywhere.

So I have been reading and I think that this question is saying that these two functions are mappings like, $$f,g:=(\mathbb{R},\mathcal{L}) \stackrel{\lambda^{\ast}}{\to} (\mathbb{R},\mathcal{B}(\mathbb{R})).$$

I started my argument by saying:


If $f,g$ are $\lambda$-a.e. on $\mathbb{R}$ then $\exists$ a set $E:=\{x \in \mathbb{R} : f \neq g\}$. But the claim incorporates that $f=g$ everywhere and so there can't exist a set on $\mathbb{R}$ that has Lebesgue outer measure zero, i.e. all sets on $\mathbb{R}$ must have positive Lebesgue measure....


I think the reason there is no set of zero Lebesgue measure is because the preimage of all similar type sets in the Borel algebra, $\mathcal{B}(\mathbb{R})$, is included in a similar set in the algebra of Lebesgue measurable sets $\mathcal{L}$. But, aren't there Lebesgue measurable sets of zero Lebesgue measure - the countable $\mathbb{N}$ for instance? Or does it not count because the preimage is coming from a Borel algebra, even though a Borel set has measure zero too?

Thanks for any help in clearing this up for me.

$\endgroup$
5
  • 2
    $\begingroup$ You are making claims that you probably don't want to make: «there can't exist a set on R that has Lebesgue outer measure zero», «all sets on R must have positive Lebesgue measure». They are obviously false. For one thing, every empty subset of $R$ has outer Lebesgue measure zero and does not have positive Lebesgue measure... $\endgroup$ Commented Oct 26, 2011 at 22:34
  • $\begingroup$ There are heaps of sets with Lebesgue outer measure zero. Single points have outer measure zero. This question is asking to show your set $E$ is empty, and the empty set has outer measure zero as well. What you probably do want is that every non-empty open subset of the reals has positive outer measure. $\endgroup$ Commented Oct 26, 2011 at 22:36
  • $\begingroup$ @nate: sets of Lebesgue measure zero do exist! $\endgroup$ Commented Oct 26, 2011 at 22:38
  • $\begingroup$ The set $E$ exists, but it's the empty set, that's what the question wants you to show. $\endgroup$ Commented Oct 26, 2011 at 22:38
  • $\begingroup$ Thank you all for your help - I'll need a few minutes to think about all this $\endgroup$
    – nate
    Commented Oct 26, 2011 at 22:44

3 Answers 3

6
$\begingroup$

If $f$ and $g$ are a.e. equal, show that the set $\{x\in\mathbb R:f(x)= g(x)\}$ is dense. The using the fact that $f$ and $g$ are continuous, show that $f=g$ on all of $\mathbb R$.

Altenratively, show that if $f\neq g$, then $f$ and $g$ differ on a whole open interval, so that the set $\{x\in\mathbb R:f(x)= g(x)\}$ is not almost $\mathbb R$.

$\endgroup$
3
  • $\begingroup$ I was wondering, for $\mathbb{R}$, what really is the difference between $\lambda$-a.e and dense? $E$ is dense in $\mathbb{R}$ if either every point in $\mathbb{R}$ is in $E$ or is arbitrarily close to a point in $E$. This seems to exclude only singletons - pretty much just like $\lambda$-a.e. What is the difference in this case? Also, the continuity of $f$ and $g$ assures there is a limit at every $x$ in their domain so then if $f$ and $g$ agree everywhere outside of $E$, then this assures that, as their in only 1 limit at $x$ outside $E$, then $f=g$ outside of $E$ too. Not sure about it all $\endgroup$
    – nate
    Commented Oct 27, 2011 at 0:09
  • $\begingroup$ $\mathbb Q$ is dense and most certainly not almost everywhere present! You really need to consider more examples of sets :) $\endgroup$ Commented Oct 27, 2011 at 0:13
  • $\begingroup$ I guess mainly I was unclear as to why denseness was needed to show $f=g$ everywhere, when all that I see needed is continuity ($f=g$ a.e. already) $\endgroup$
    – nate
    Commented Oct 27, 2011 at 1:17
6
$\begingroup$

Consider $f-g$, which is zero almost everywhere and continuous. If $f-g$ is non-zero at some point $x$, then by continuity it is also non-zero in some open ball around $x$. Since this open ball has positive measure, it contradicts the assumption that $f-g$ is zero almost everywhere.

$\endgroup$
2
$\begingroup$

If $f\neq g$ everywhere, then there exists $x_0$ such that $f(x_0)>g(x_0)$ (for example). By continuity of $f,g$ it follows that $f(x)>g(x)$ on a neighborhood $(x_0-\varepsilon,x_0+\varepsilon)$ of $x_0$. and that means that the set $[f\neq g]$ has measure at least $2\varepsilon>0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .